Download Solution of ECE 300 Test 2 F08

Solution of ECE 300 Test 2 F08
1.
Find the numerical value of I in the circuit below.
I
R = 5Ω
2
3
1
2.
V1
R2 + R3 || R4
=
4
R = 6Ω
V = 10V
I=
R = 3Ω
10V
10V
10V
=
=
= 1.429A
5 + 6 || 3 5 + 2 7
Find the numerical value of V in the circuit below.
R4 = 2Ω
R2 = 3Ω
R = 6Ω
V
3
R = 1Ω
1
I = 1A
1
Voltage across the 1 resistor (positive on top) is 1V. Voltage across 3 resistor (positive on the left) is
3V. Therefore the voltage across the 6 resistor (positive on top) is 4V and the current through the 6
resistor is 2/3 A. So the current through the 2 resistor (flowing to the right) is 5/3 A and the voltage
across it (positive on the left) is 10/3V. Then the voltage V must be 4V + 10/3V or 22/3V or 7.333 V.
3.
Find the numerical values of the three currents indicated on the circuit diagram below.
R4 = 4Ω
8Ω
I
12Ω
s
R3 = 10Ω
5V
I
8
Ix
Current through the 10 resistor is 1/2 A flowing downward. Therefore the current through the 4
resistor is 1/2 A flowing to the right and the voltage across the 4 resistor is 2V (positive on the left). So
the voltage across the current source and the 8 and 12 resistors is 7V (positive on top). Therefore the
current I8 is 7/8A and the current through the 12 resistor is 7/12A flowing downward. Therefore I s is
the sum of 7/8, 7/12 and 1/2 or 1.958A. I x must be the same as the sum of the currents flowing through
the 8 and 12 resistors or 1.458A.
4.
Find the numerical value of V80 in the circuit below.
200 mA
R1 = 100kΩ
R2 = 50kΩ
R3 = 30kΩ
R = 80kΩ
V
4
80
The 200 mA current splits between the two paths. The current through the right-hand path is (using
current splitting)
200mA 30k
= 37.5mA
30k + 50k + 80k
Therefore the voltage across the 80k resistor is 37.5 mA 80 k = 3000 V
5.
Find the numerical value of V in the circuit below.
3V
500Ω
2kΩ
8kΩ
7V
10V
3kΩ
V
The three voltage sources in series can be replaced by a single voltage source in the position, and with the
polarity of the 7V source, of –6V. The voltage V is then (using voltage division)
3000
V = 6V = 4 / 3V = 1.333V
3000 + 2000 + 500 + 8000 Solution of ECE 300 Test 2 F08
1.
Find the numerical value of I in the circuit below.
I
R = 7Ω
2
3
1
2.
V1
R2 + R3 || R4
=
4
R = 6Ω
V = 10V
I=
R = 3Ω
10V
10V
10V
=
=
= 1.111A
7 + 6 || 3 7 + 2 9
Find the numerical value of V in the circuit below.
R4 = 2Ω
R2 = 2Ω
R = 6Ω
V
3
R = 1Ω
1
I = 1A
1
Voltage across the 1 resistor (positive on top) is 1V. Voltage across 2 resistor (positive on the left) is
2V. Therefore the voltage across the 6 resistor (positive on top) is 3V and the current through the 6
resistor is 1/2 A. So the current through the 2 resistor (flowing to the right) is 3/2 A and the voltage
across it (positive on the left) is 3V. Then the voltage V must be 3V + 3V or 6 V.
3.
Find the numerical values of the three currents indicated on the circuit diagram below.
R4 = 8Ω
8Ω
I
12Ω
s
R3 = 10Ω
5V
I
8
Ix
Current through the 10 resistor is 1/2 A flowing downward. Therefore the current through the 8
resistor is 1/2 A flowing to the right and the voltage across the 8 resistor is 4V (positive on the left). So
the voltage across the current source and the 8 and 12 resistors is 9V (positive on top). Therefore the
current I8 is 9/8A and the current through the 12 resistor is 3/4A flowing downward. Therefore I s is
the sum of 9/8, 3/4 and 1/2 or 2.375A. I x must be the same as the sum of the currents flowing through
the 8 and 12 resistors or 1.875A.
4.
Find the numerical value of V60 in the circuit below.
200 mA
R1 = 100kΩ
R2 = 50kΩ
R3 = 30kΩ
R = 60kΩ
V
4
60
The 200 mA current splits between the two paths. The current through the right-hand path is (using
current splitting)
200mA 30k
= 42.86mA
30k + 50k + 60k
Therefore the voltage across the 60k resistor is 42.86 mA 60 k = 2571.6 V
5.
Find the numerical value of V in the circuit below.
3V
500Ω
2kΩ
4kΩ
7V
10V
3kΩ
V
The three voltage sources in series can be replaced by a single voltage source in the position, and with the
polarity of the 7V source, of –6V. The voltage V is then (using voltage division)
3000
V = 6V = 1.895V
3000 + 2000 + 500 + 4000